{"id":2099,"date":"2022-02-22T13:08:50","date_gmt":"2022-02-22T04:08:50","guid":{"rendered":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/?page_id=2099"},"modified":"2024-04-17T10:23:39","modified_gmt":"2024-04-17T01:23:39","slug":"%e5%8f%82%e8%80%83%ef%bc%9a%e4%b8%89%e8%a7%92%e9%96%a2%e6%95%b0%e3%81%ae%e5%be%ae%e5%88%86%e3%81%ae%e3%82%82%e3%81%86%e4%b8%80%e3%81%a4%e3%81%ae%e5%b0%8e%e5%87%ba%e6%b3%95","status":"publish","type":"page","link":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/%e7%90%86%e5%b7%a5%e7%b3%bb%e3%81%ae%e6%95%b0%e5%ad%a6b\/%e4%b8%89%e8%a7%92%e9%96%a2%e6%95%b0%e3%81%ae%e5%be%ae%e5%88%86\/%e5%8f%82%e8%80%83%ef%bc%9a%e4%b8%89%e8%a7%92%e9%96%a2%e6%95%b0%e3%81%ae%e5%be%ae%e5%88%86%e3%81%ae%e3%82%82%e3%81%86%e4%b8%80%e3%81%a4%e3%81%ae%e5%b0%8e%e5%87%ba%e6%b3%95\/","title":{"rendered":"\u53c2\u8003\uff1a\u4e09\u89d2\u95a2\u6570\u306e\u5fae\u5206\u306e\u3082\u3046\u4e00\u3064\u306e\u5c0e\u51fa\u6cd5"},"content":{"rendered":"

<\/p>\n

\u4e09\u89d2\u95a2\u6570\u306e\u5fae\u5206\u3092\u793a\u3059\u305f\u3081\u306b\u5fc5\u8981\u306a\u306e\u306f\uff0c\u6b21\u306e3\u3064\u3002\u3053\u308c\u3089\u306f\u9ad8\u6821\u306e\u6570\u5b66\u3067\u7fd2\u3063\u305f\u306f\u305a\u3067\u3059\u304b\u3089\u8a3c\u660e\u306f\u7565\u3002<\/p>\n

1. \\(\\quad\\) \u4e09\u89d2\u95a2\u6570\u306e\u6975\u9650\u516c\u5f0f \u00a0 \\(\\quad\\displaystyle \\lim_{x \\rightarrow 0} \\frac{\\sin x}{x} = 1 \\)<\/p>\n

2.\u00a0 \\(\\displaystyle \\quad \\sin\\alpha -\\sin\\beta = 2 \\cos\\frac{\\alpha+\\beta}{2} \\sin\\frac{\\alpha-\\beta}{2} \\)<\/p>\n

3.\u00a0\u00a0\\(\\displaystyle \\quad \\cos\\alpha -\\cos\\beta = -2 \\sin\\frac{\\alpha+\\beta}{2} \\sin\\frac{\\alpha-\\beta}{2} \\)<\/p>\n

\u5c0e\u95a2\u6570\u306e\u5b9a\u7fa9\u304b\u3089\uff0c<\/p>\n

\\begin{eqnarray}
\n(\\sin x)’ &=& \\lim_{h \\rightarrow 0} \\frac{\\sin(x + h) -\\sin x}{h} \\\\
\n&=& \\lim_{h \\rightarrow 0} \\frac{1}{h} \\cdot 2 \\cos\\frac{x + h + x}{2} \\sin\\frac{x + h -x}{2}\\\\
\n&=& \\cos x\\cdot \\lim_{\\frac{h}{2} \\rightarrow 0} \\frac{\\sin\\frac{h}{2}}{\\frac{h}{2}} \\\\
\n&=& \\cos x
\n\\end{eqnarray}<\/p>\n

\\( (\\cos x)’\\) \u3082\u540c\u69d8\u306b\u8a08\u7b97\u3067\u304d\u308b\u3002\\( (\\tan x)’ \\) \u306b\u3064\u3044\u3066\u306f\uff0c\u5fae\u5206\u6cd5\u306e\u516c\u5f0f 5. \u3088\u308a
\n$$ (\\tan x)’ = \\left\\{ \\frac{\\sin x}{\\cos x} \\right\\}’\u00a0 = \\frac{(\\sin x)’ \\cos x -\\sin x (\\cos x)’}{\\cos^2 x} = \\cdots$$
\n\u3068\u3057\u3066\uff0c\\( \\cos^2 x + \\sin^2 x = 1\\) \u3092\u4f7f\u3046\u306e\u3067\u3042\u3063\u305f\u3002<\/p>\n","protected":false},"excerpt":{"rendered":"","protected":false},"author":33,"featured_media":0,"parent":2096,"menu_order":1,"comment_status":"closed","ping_status":"closed","template":"","meta":{"inline_featured_image":false,"footnotes":""},"class_list":["post-2099","page","type-page","status-publish","hentry","nodate","item-wrap"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t